A lower bound for the length of a partial transversal in a Latin square

An array is row-Latin if no symbol is repeated within any row. An array is Latin if it and its transpose are both row-Latin. A transversal in an n × n array is a selection of n different symbols from different rows and different columns. We prove that every n × n Latin array containing at least (2 − √ 2)n 2 distinct symbols has a transversal. Also, every n × n row-Latin array containing at least 1 4 (5 − √ 5)n 2 distinct symbols has a transversal. Finally, we show by computation that every Latin array of order 7 has a transversal, and we describe all smaller Latin arrays that have no transversal.

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“…Indeed, any linear upper bound on

ℓ (n)

would imply the existence of a constant c such that every Latin square of order n has a partial transversal of length

n - c

. The best result to date is that every Latin square has a partial transversal of length

n - O ({prefixlog}^{2} n)

.…”

Section: Introductionmentioning

confidence: 99%

Transversals in Latin Arrays with Many Distinct Symbols

Best

Hendrey

Wanless

et al. 2017

J of Combinatorial Designs

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“…Now our desired matching corresponds to a transversal. Hatami and Shor [9] proved that if K m,m is a union of m 1-factors F 1 , F 2 , . .…”

Section: Introduction and Notationmentioning

confidence: 99%

Orthogonal matchings revisited

Wang

2015

Discrete Mathematics

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a b s t r a c tLet G be a graph on n vertices, which is an edge-disjoint union of ms-factors, that is, s regular spanning subgraphs. Alspach first posed the problem that if there exists a matching M of m edges with exactly one edge from each 2-factor. Such a matching is called orthogonal because of applications in design theory. For s = 2, so far the best known result is due to Stong in 2002, which states that if n ≥ 3m−2, then there is an orthogonal matching. Anstee and Caccetta also asked if there is a matching M of m edges with exactly one edge from each s-factor? They answered yes for s ≥ 3. In this paper, we get a better bound and prove that if s = 2 and n ≥ 2 √ 2m+4.5 (note that 2 √ 2 ≤ 2.825), then there is an orthogonal matching. We also prove that if s = 1 and n ≥ 3.2m − 1, then there is an orthogonal matching, which improves the previous bound (3.79m).

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“…The best known lower bound is due to Hatami and Shor , who showed that every Latin square of order n has a partial transversal of size